TITLE: Eulerian number identity
QUESTION [16 upvotes]: The Eulerian number $A(n,m)$ is defined as the number of permutations $\sigma \in S_n$ having precisely $m$ descents, i.e. indices $i$ such that $\sigma(i)>\sigma(i+1)$. 
The wikipedia entry on these numbers mentions one identity for which I cannot locate a proof, and I would be grateful for a reference. Namely
$$\sum_{m=0}^{n-1} (-1)^m \frac{A(n,m)}{\left(n\atop m\right)}=(n+1) B_n\ \ \forall n\ge 2.$$ Here $\{B_n\}$ are the Bernoulli numbers.
(I do understand the two related identities immediately preceding this one. The first of these can be found in many places, and the second is easy.)

REPLY [3 votes]: A scent that leads to interesting creatures and vistas (and several different derivations of this Eulerian identity) is a blend of
$h(t) = e^{t}-1 \; \; \; $      and $\; \; \; h^{(-1)}(t) = \ln(1+t)$.
Follow this and you encounter paths blazed by Bernoulli, Stirling, Euler, and even the ancient wizard Archimedes with the help of the bold Blissard.
First denote the Bernoulli numbers $b_n$ and polynomials $Ber_n(x) = (b.+x)^n = \sum_{k=0}^n \binom{n}{k} \; b_k \; x^{n-k} ,$ then
$$ \; t\;D_t \ln(e^{-t}-1) =  \frac{t}{e^t-1}  = e^{b.t} .$$
(Notice this relates the zeros of $e^{-t}-1$ to the Riemann zeta fct. and the Bernoullis, as Riemann noted in his famous paper.)
Transform this with the inverse $t = -\ln(1+u)$ to get
$$\frac{\ln(1+u)}{u} (1+u) = e^{b.(-\ln(1+u))}= (1+u)^{-b.},$$
or
$$\frac{\ln(1+u)}{u} = e^{a.u} = (1+u)^{-b.-1},$$
where $a_n= \frac{(-1)^nn!}{n+1} .$
Then since the Bernoulli polynomials, as do all Appells, have the shift property $(Ber.(x)+y)^n= (b.+x+y)^n = Ber_n(x+y)$, the binomial expansion gives
$$\frac{a_n}{n!} = \binom{-b.-1}{n} = (-1)^n \binom{b.+n}{n}$$
$$ =\binom{-Ber.(1)}{n}=(-1)^n\binom{Ber.(n)}{n}$$
$$ = \binom{Ber.(0)}{n} = \binom{b.}{n},$$
or, equivalently,
$$   a_n = ST1_n(-Ber_.(1)) = (-1)^n  ST1_n(Ber.(n)) = ST1_n(b.). $$
where $ST1_n(x)$ are the Stirling polynomials of the first kind.
(The relation $(-1)^n Ber_n(1) = Ber_n(0)=b_n$ follows from the e.g.f. for the Bernoulli polynomials that follow from their natural operational definition $f(Ber.(x+1)) - f(Ber.(x)) = \frac{d}{dx} f(x)$. The sign only makes a difference for $n=1$ since all other odd order $b_n$ equal zero.)
Then, using the umbral compositional inverse relation between the Stirling polynomials of the first and second kinds
$$ST1_n(ST2.(x))=x^n=ST2_n(ST1.(x)),$$
we have
$$ST2_n(a.) = ST2_n(ST1.(b.)) = (b.)^n = b_n,$$
and the less elegant, trickier relation
$$ST2_n(-a.) = \sum_{k=0}^n ST2_{n,k} ST1_k(B.(k)) = \sum_{k=0}^n ST2_{n,k} \sum_{k=0}^m ST1_{k,m}B_m(k)).$$
Now take one of the Archimedean solids--the truncated octahedron--generalize it to all dimensions, and call the resulting simple convex polytopes the permutahedra (or permutohedra). Then we can relate the Stirling numbers of the second kind, $ST2_{n,k}$, and, consequently, the first kind, $ST1_{n.k}$, to the Eulerian numbers, for the h-polynomials of the permutahedra are the Eulerian polynomials $E_n(x) = \sum_{k=0}^n E_{n.k} \ x^k$, and the corresponding face polynomials $F_n(x)$ are related via $F_n(x) = E_n(1+x)$ (see footnote for examples). Now the reverse face polynomials are
$$RF_n(x) = x^n F_n(1/x) = \sum_{k=-0}^n ST2_{n,k}\;k!\; x^k= x^n E_n(1+1/x) ,$$
or, changing signs,
$$\sum_{k=0}^n ST2_{n,k}(-1)^k \;k!\; x^k = \sum _{k=0}^n E_{n,k} \sum_{m=0}^k (-1)^{n+m} \binom{k}{m} x^{n-m}.$$
Integrating from $0$ to $1$,
$$ST2_n(a.) = \sum _{k=0}^n E_{n,k} \sum_{m=0}^k \binom{k}{m} (-1)^{n+m}\frac{1}{n-m+1}.$$
Focusing on the last summation,
$$ \frac{1}{(n+1)!}  = D_{x=1}^{j-n-1} \frac{x^{j}}{j!} = \int_{0}^1 \frac{u^{n-j}}{(n-j)!}\frac{(1-u)^j}{j!} du$$
$$ = \frac{(-1)^{j+n}}{(n-j)!j!} \int_{0}^1 (-u)^n  \sum_{m=0}^{j} (-u)^{-m} \binom{j}{m}du$$
$$ = \frac{(-1)^{j+n}}{(n-j)!j!} \sum_{m=0}^j \frac{(-1)^{n-m}}{n-m+1} \binom{j}{m},$$
giving
$$ \sum_{m=0}^j \frac{(-1)^{n-m}}{n-m+1} \binom{j}{m} = (-1)^{j+n}\frac{(n-j)!j!}{(n+1)!} \; .$$
So equating the two expressions for $ST2_n(a.)$, we arrive at, for $n \geq 0$,
$$ b_n = \sum_{j=0}^{n} E_{n,j} \sum_{m=0}^j \frac{(-1)^{n-m}}{n-m+1} \binom{j}{m}= \sum_{j=0}^{n} (-1)^{j+n} E_{n,j} \frac{(n-j)!j!}{(n+1)!}.$$
These are the infants mimicking the adults--
$$Ber_n(x) = ST2_n(q.(ST1.(x))),$$
with  $$q_n(x) = (q.(0)+x)^n = (a. + x)^n.$$
If we go deeper into the forest, we encounter Worpitzky, Dobinski, Kummer, Laguerre, Lah, the positroids and their consorts the stellahedra, among others,
but that's another episode.
(Only the Shadow knows what magic dwells in the heart of forests.)
(But like John Mayer in his song) one more thing. Returning to the relation between the f-polynomials and the Eulerians, we see that
$$Bell_n(x) = ST2_n(x) = \sum_{j=0}^n ST2_{n,j} \; x^j = \sum_{j=0}^n \frac{x^j}{j!}\sum_{k=0}^n E_{n,k} \binom{k}{n-j}$$
$$ = \sum_{k=0}^n E_{n,k} \sum_{m=0}^n  \binom{m}{n} \frac{x^{j-m}}{(j-m)!} = \sum_{k=0}^n E_{n,k} L_n^{k-n}(-x),$$
introducing the associated Laguerre polynomials
$$L_n^{\alpha}(-x) = \sum_{k=0}^n\; \binom{n+\alpha}{k+\alpha}\;\frac{x^k}{k!}.$$
(Befitting the entire reciprocal Euler gamma function, I use the convention $1/(-n)! =0 $ for $n=1,2,3,...$ .)
Invoking the Chu-Vandermonde identities (alternatively see this MO-Q),
$$ L_n^{k-n}(-St1.(x)) = \sum_{j=0}^n \binom{k}{n-j} \frac{ST1_{j}(x)}{j!} $$
$$= \sum_{j=0}^n  \binom{k}{n-j} \binom{x}{j} =  \binom{x+k}{n},$$
implying after composing with $ST2.(x) = Bell.(x)$ that
$$ L_n^{k-n}(-x)=   \binom{Bell.(x)+k}{n},$$
giving the Bell morphed Worpitzky identity,
$$ Bell_n(x)= \sum_{k=0}^n E_{n,k} \binom{Bell.(x)+k}{n}.$$
By umbrally composing this with $ST1.(x)$, we morph back to the canonical WI,
$$ x^n= \sum_{k=0}^n E_{n,k} \binom{x+k}{n},$$
and, umbrally morphing again,
$$ Ber_n(x)= \sum_{k=0}^n E_{n,k} \binom{Ber.(x)+k}{n} = \sum_{k=0}^n E_{n,k} \binom{Ber.(x+k)}{n}.$$
Evaluating at $x=0$ and comparing with the previous identities,
$$ \binom{Ber.(m)}{n} =  \binom{b.+m}{n} = (-1)^{m+n} \frac{(n-m)!m!}{(n+1)!}.$$
(Edit, Dec 12, 2020: This identity is valid only for $m \leq n = 0,1,2,...$. For quick checks, $ D_x \binom{x}{n+1} = \binom{Ber.(x)}{n}$. This and the WI imply $\sum_{k=0}^{n} E_{n,k} \binom{Ber.(k)}{n-1} = 0$, consistent with palindromic symmetry of the Eulerians and skew (signed) palindromic symmetry of the umbral binomials of the Bernoulli numbers. Actually, this last identity allows the extension of the validity of the chief identity to $m = n$.)
(Again, there is a deeper story here based on a dance between the $h(t)= e^t-1$ and its consort $h^{(-1)}(t)= \ln(1+t)$ and similar couples.)

For nailing down indices, definitions, and spot checking these identities and any new ones someone may discover:
$Ber_n(x) = (b. +x)^n = \sum_{k=0}^n \binom{n}{k} \; b_n \;x^{n-k}$
$$e^{B.(x)t} = \frac{t}{e^t-1} e^{x t} = e^{b.t} e^{xt}=e^{(b.+x)t}$$

$b_0 = 1\;\;\;\;\;\;\;\;\;\;$ $B_0(x)=1$
$b_1 = -1/2\;\;\;\;\;\;\;\;\;\;$ $B_1(x)= \frac{-1+2x}{2}$
$b_2 = 1/6\;\;\;\;\;\;\;\;\;\;$ $B_2(x)= \frac{1-6x+6x^2}{6}$
$b_3 = 0\;\;\;\;\;\;\;\;\;\;$ $B_3(x)= \frac{x-3x^2+2x^3}{2}$

$ST1_n(x) = \sum_{k=0}^n ST1_{n,k}\; x^k \;\;\;\;\;$    (a.k.a. the falling factorials)
$$e^{ST1.(x) t} = e^{x\ln(1+t)} = (1+t)^x$$

$ST1_0(x) =1 $
$ST1_1(x) = x$
$ST1_2(x) = -x+x^2$
$ST1_3(x) = 2x-3x^2+x^3$

$ST2_n(x) = \sum_{k=0}^n ST2_{n,k}\; x^k \;\;\;\;\; $    (a.k.a. the Touchard / Bell / Exponential polynomials)
$$e^{ST2.(x) t} = e^{x(e^t-1)}$$

$ST2_0(x) =1 $
$ST2_1(x) = x$
$ST2_2(x) = x+x^2$
$ST2_3(x) = x+3x^2+x^3$

$RF_n(x) =  \sum_{k=0}^n RF_{n,k}\; x^k$ $$e^{RF.(x)t} = \frac{1}{1+x-xe^{t}}$$

$RF_0(x) =1 $
$RF_1(x) = x$
$RF_2(x) = x+2x^2$
$RF_3(x) = x+6x^2+6x^3$

$F_n(x) = \sum_{k=0}^n F_{n,k}\; x^k$ $$e^{F.(x)t} = \frac{x}{1+x-e^{tx}}$$

$F_0(x) =1 $
$F_1(x) = 1$
$F_2(x) = 2+x$
$F_3(x) = 6+6x+x^2$

$E_n(x) =  \sum_{k=0}^n E_{n,k}\; x^k$ $$e^{E.(x)t} = \frac{x-1}{x-e^{(x-1)t}}$$

$E_0(x) =1 $
$E_1(x) = 1$
$E_2(x) = 1+x$
$E_3(x) = 1+4x+x^2$